Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}-8x-3y &= 3 \\ 9x+3y &= -4\end{align*}$
Explanation: Begin by moving the $y$ -term in the second equation to the right side of the equation. $9x = -3y-4$ Divide both sides by $9$ to isolate $x$ $x = {-\dfrac{1}{3}y - \dfrac{4}{9}}$ Substitute this expression for $x$ in the first equation. $-8({-\dfrac{1}{3}y - \dfrac{4}{9}}) - 3y = 3$ $\dfrac{8}{3}y + \dfrac{32}{9} - 3y = 3$ Simplify by combining terms, then solve for $y$ $-\dfrac{1}{3}y + \dfrac{32}{9} = 3$ $-\dfrac{1}{3}y = -\dfrac{5}{9}$ $y = \dfrac{5}{3}$ Substitute $\dfrac{5}{3}$ for $y$ in the top equation. $-8x-3( \dfrac{5}{3}) = 3$ $-8x-5 = 3$ $-8x = 8$ $x = -1$ The solution is $\enspace x = -1, \enspace y = \dfrac{5}{3}$.